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GR - Lie Derivative of metric - Killing Equation

  1. Jul 18, 2017 #1
    1. The problem statement, all variables and given/known data

    Question attached. hell.png

    2. Relevant equations


    3. The attempt at a solution


    I'm not really sure how to work with what is given in the quesiton without introducing my knowledge on lie derivatives.

    We have: ##(L_ug)_{uv} = U^{\alpha}\nabla_{\alpha}g_{uv}+g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}##

    Where ##L_u## denotes the lie derivative in the direction of ##u##

    Where the first term vanishes on the assumption of a levi-civita connection by the fundamental theorem of riemmanian geometry. ##U^{\alpha}\nabla_{\alpha}g_{uv}= U^{\alpha}(\partial_{\alpha}g_{uv} + \Gamma^{c}_{\alpha v} g_{uc} + \Gamma^{c}_{\alpha u} g_{vc}) ##

    And so the fact that ##w^u \partial_u g_{ab} =0 \implies \partial_t g_{ab} =0 ## vanishes here, implies that the connection terms vanish (here I need an arguement that they vanish individually, and that the sum can not vanish) I can then conclude that this implies we are working in Minkowski space-time.

    And I can then use my the following knowledge on Lie derivatives to obtain an answer:

    ##w^u \partial_u g_{ab} =0 \implies \partial_t g_{ab} =0 ## ; i.e. the metric has no ##t## dependence so I know that this means that the Lie derivative (1) vanishes:

    And then I have

    ##(L_ug)_{uv} =0= g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}=\nabla_vU_u+\nabla_u U_v##

    ##\implies k=-1 ##

    However the question makes no reference to the requirement of needing Lie derivative, so I'm not too sure about what I've done here,

    Can I start from first principles more? Any hint appreciated.

    Thanks in advance.
     
  2. jcsd
  3. Jul 18, 2017 #2
    Here's what I think is one way. Consider [tex] \nabla_\mu W_\nu = \partial_\mu W_\nu -\Gamma^s_{\nu \mu} W_s = \partial_\mu W_\nu -\tfrac12 W_s g^{s \sigma} ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu} - \partial_\sigma g_{\nu \mu}) = \partial_\mu W_\nu -\tfrac12 W^\sigma ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu} - \partial_\sigma g_{\nu \mu}) = \partial_\mu W_\nu -\tfrac12 W^\sigma ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu}) [/tex]
    where the last term vanishes by assumption. Now since [itex] W^\mu = (1,0,0,0) [/itex], [itex] W_\mu = g_{\nu \mu} W^\nu = g_{0 \mu}[/itex]. Therefore the equation can be rewritten as [tex] \partial_\mu W_\nu -\tfrac12 W^\sigma ( \partial_\mu g_{\sigma \nu } + \partial_\nu g_{\sigma \mu}) = \partial_\mu g_{0 \nu} - \tfrac12 ( \partial_\mu g_{0 \nu} + \partial_\nu g_{0 \mu}) = \tfrac12 \partial_\mu g_{0 \nu} - \tfrac12 \partial_\nu g_{0 \mu}. [/tex] It is easy to check that [itex] \nabla_\nu W_\mu = -(\tfrac12 \partial_\mu g_{0 \nu} - \tfrac12 \partial_\nu g_{0 \mu}) [/itex], so [itex]k=-1[/itex] as expected.
     
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